Bit Manipulation

Bitwise NOT

~val

NOT 0111 (decimal 7)
= 1000 (decimal 8)

~val is also equivalent to val != -1, since -1 is the only one number whose bitwise NOT is 0.

So if val is some non-negative number, the following:

for (int i = val; ~i; --i)

is equivalent to

for (int i = val; i >= 0; --i)

Set the ith bit

mask |= 1 << i

Unset the ith bit

mask &= ~(1 << i)

Toggle the ith bit

mask ^= 1 << i

Check if ith bit is set

mask & (1 << i)
// Or
mask >> i & 1 // less typing

Set the ith bit with x

mask = (mask & ~(1 << i)) | (x << i)

Get the lowest bit

function lowbit(int x) {
    return x & -x;
}

Example:

        12 = (01100)2
       -12 = (10100)2
12 & (-12) = (00100)2

Count bit 1s

__builtin_popcount(n);

Note that the pop in this function means population.

Count trailing zeros

__builtin_ctz(n);

Count leading zeros

__builtin_clz(n);

Check if n is power of 2

__builtin_popcount(n) == 1
// or
(n & (n - 1)) == 0 // & has lower precedence than ==. Must add the parenthesis.

Traverse all the subsets

for (int mask = 0; mask < (1 << N); ++mask)

Complement of a subset

// Assume `sub` is a subset of `mask`
int complement = mask ^ sub;
// Or
int complement = mask - sub;

Traverse subsets of a set mask

for (int sub = mask; sub; sub = (sub - 1) & mask) {
    // `sub` is a non-empty subset of `mask`
}

Sample output:

// mask = 0b1011
1011
1010
1001
1000
0011
0010
0001

Or the other direction:

for (int other = mask; other; other = (other - 1) & mask) {
    int sub = mask - other;
    // `other` is a non-empty subset of `mask`. `sub` is the complement of `other`.
}

Sample output:

0000
0001
0010
0011
1000
1001
1010

Given N elements, traverse subsets of size K (Gosper's Hack)

// Time Complexity: O(C(N, K))
int sub = (1 << k) - 1;
while (sub < (1 << N)) {
    // Access `sub` here. `sub` has `K` elements and is a subset of `N` elements.
    int c = sub & - sub, r = sub + c;
    sub = (((r ^ sub) >> 2) / c) | r;
}

Traverse subsets of subsets

Given an array of N elements, traverse all its subsets. For each of the subsets, traverse its all non-empty subsets as well.

// Time: O(3^N)
for (int mask = 0; mask < (1 << N); ++mask) {
    // `mask` is a subset of `N` elements
    for (int sub = mask; sub; sub = (sub - 1) & mask) {
        // `sub` is a subset of `mask`.
    }
}

Note that the time complexity is O(3^N) instead of O(2^N * 2^N) = O(4^N).

Math proof:

For a subset mask with K bits of 1s, there are 2^K subsets of it.

For N elements, there are C(N, K) subsets with K bits of 1s.

So the total number is SUM( C(N, K) * 2^K | 0 <= K <= N ).

Since (1 + x)^N = C(N, 0) * x^0 + C(N, 1) * x^1 + ... + C(N, N) * x^N, let x = 2, we have 3^N = SUM( C(N, K) * 2^K | 0 <= K <= N ).

Intuitive proof:

For each bit index i (0 <= i < N), the i-th bits of mask and sub must be one of 11, 10, 00, and can't be 01. So each bit has 3 possibilities, the total complexity is O(3^N).

Calculate subset sums

// Time: O(N * 2^N)
for (int mask = 0; mask < (1 << N); ++mask) {
    for (int i = 0; i < N; ++i) {
        if (mask >> i & 1) sum[mask] += A[i];
    }
}

Or

// DP solution
// Time: O(2^N)
for (int mask = 1; mask < (1 << N); ++mask) {
    sum[mask] = sum[mask - lowbit(mask)] + A[__builtin_ctz(lowbit(mask))];
}

where lowbit(x) = x & -x, and __builtin_ctz(x) counts the trailing zeros in x. Since lowbit(mask) is a power of two, __builtin_ctz(lowbit(mask)) == log2(lowbit(mask)).

Generate logs array

logs[n] is floor(log2(n)).

logs[1] = 0
logs[2] = 1
logs[3] = 1
logs[4] = 2
logs[5] = 2
logs[6] = 2
logs[7] = 2
logs[8] = 3
...

// Time: O(2^N)
for (int mask = 2; mask < (1 << N); ++mask) {
    logs[mask] = logs[mask >> 1] + 1;
}

Reference

• https://oi-wiki.org/math/bit/